Optimal. Leaf size=27 \[ \frac {1}{2} \log \left (x^2\right ) \log \left (x+\log \left (\left (1-\frac {e^2}{2}+e^x+x\right )^2\right )\right ) \]
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Rubi [F] time = 2.71, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {\left (6 x-e^2 x+6 e^x x+2 x^2\right ) \log \left (x^2\right )+\left (4 x-2 e^2 x+4 e^x x+4 x^2+\left (4-2 e^2+4 e^x+4 x\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right ) \log \left (x+\log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right )}{4 x^2-2 e^2 x^2+4 e^x x^2+4 x^3+\left (4 x-2 e^2 x+4 e^x x+4 x^2\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (6 x-e^2 x+6 e^x x+2 x^2\right ) \log \left (x^2\right )+\left (4 x-2 e^2 x+4 e^x x+4 x^2+\left (4-2 e^2+4 e^x+4 x\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right ) \log \left (x+\log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right )}{4 e^x x^2+\left (4-2 e^2\right ) x^2+4 x^3+\left (4 x-2 e^2 x+4 e^x x+4 x^2\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )} \, dx\\ &=\int \left (\frac {\left (6 e^x+6 \left (1-\frac {e^2}{6}\right )+2 x\right ) \log \left (x^2\right )}{2 \left (2 e^x+2 \left (1-\frac {e^2}{2}\right )+2 x\right ) \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )}+\frac {\log \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )}{x}\right ) \, dx\\ &=\frac {1}{2} \int \frac {\left (6 e^x+6 \left (1-\frac {e^2}{6}\right )+2 x\right ) \log \left (x^2\right )}{\left (2 e^x+2 \left (1-\frac {e^2}{2}\right )+2 x\right ) \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )} \, dx+\int \frac {\log \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )}{x} \, dx\\ &=\frac {1}{2} \int \left (\frac {3 \log \left (x^2\right )}{x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )}+\frac {2 \left (e^2-2 x\right ) \log \left (x^2\right )}{\left (2 e^x+2 \left (1-\frac {e^2}{2}\right )+2 x\right ) \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )}\right ) \, dx+\int \frac {\log \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )}{x} \, dx\\ &=\frac {3}{2} \int \frac {\log \left (x^2\right )}{x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )} \, dx+\int \frac {\left (e^2-2 x\right ) \log \left (x^2\right )}{\left (2 e^x+2 \left (1-\frac {e^2}{2}\right )+2 x\right ) \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )} \, dx+\int \frac {\log \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )}{x} \, dx\\ &=\frac {3}{2} \int \frac {\log \left (x^2\right )}{x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )} \, dx+\int \left (\frac {2 x \log \left (x^2\right )}{\left (-2 e^x-2 \left (1-\frac {e^2}{2}\right )-2 x\right ) \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )}+\frac {e^2 \log \left (x^2\right )}{\left (2 e^x+2 \left (1-\frac {e^2}{2}\right )+2 x\right ) \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )}\right ) \, dx+\int \frac {\log \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )}{x} \, dx\\ &=\frac {3}{2} \int \frac {\log \left (x^2\right )}{x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )} \, dx+2 \int \frac {x \log \left (x^2\right )}{\left (-2 e^x-2 \left (1-\frac {e^2}{2}\right )-2 x\right ) \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )} \, dx+e^2 \int \frac {\log \left (x^2\right )}{\left (2 e^x+2 \left (1-\frac {e^2}{2}\right )+2 x\right ) \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )} \, dx+\int \frac {\log \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right )}{x} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 1.09, size = 33, normalized size = 1.22 \begin {gather*} \frac {1}{2} \log \left (x^2\right ) \log \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.23, size = 45, normalized size = 1.67 \begin {gather*} \frac {1}{2} \, \log \left (x^{2}\right ) \log \left (x + \log \left (x^{2} - {\left (x + 1\right )} e^{2} + {\left (2 \, x - e^{2} + 2\right )} e^{x} + 2 \, x + \frac {1}{4} \, e^{4} + e^{\left (2 \, x\right )} + 1\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 2.05, size = 247, normalized size = 9.15
method | result | size |
risch | \(\ln \relax (x ) \ln \left (-2 \ln \relax (2)+2 \ln \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )-\frac {i \pi \,\mathrm {csgn}\left (i \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )^{2}\right ) \left (-\mathrm {csgn}\left (i \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )^{2}\right )+\mathrm {csgn}\left (i \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )\right )\right )^{2}}{2}+x \right )-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (\mathrm {csgn}\left (i x \right )^{2}-2 \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )+\mathrm {csgn}\left (i x^{2}\right )^{2}\right ) \ln \left (\ln \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )-\frac {i \left (\pi \mathrm {csgn}\left (i \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )\right )^{2} \mathrm {csgn}\left (i \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )^{2}\right )-2 \pi \,\mathrm {csgn}\left (i \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )\right ) \mathrm {csgn}\left (i \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )^{2}\right )^{2}+\pi \mathrm {csgn}\left (i \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )^{2}\right )^{3}-4 i \ln \relax (2)+2 i x \right )}{4}\right )}{4}\) | \(247\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.68, size = 26, normalized size = 0.96 \begin {gather*} \log \left (x - 2 \, \log \relax (2) + 2 \, \log \left (2 \, x - e^{2} + 2 \, e^{x} + 2\right )\right ) \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.81, size = 48, normalized size = 1.78 \begin {gather*} \frac {\ln \left (x^2\right )\,\ln \left (x+\ln \left (2\,x+{\mathrm {e}}^{2\,x}+\frac {{\mathrm {e}}^4}{4}+\frac {{\mathrm {e}}^x\,\left (8\,x-4\,{\mathrm {e}}^2+8\right )}{4}+x^2-\frac {{\mathrm {e}}^2\,\left (4\,x+4\right )}{4}+1\right )\right )}{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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